Careercup - Microsoft面试题 - 5799446021406720

2014-05-12 07:17

题目链接

原题：

Given below is a tree/trie 

A 

B    c D 

e    F 

a<b<e<>>c<>d<f<>>> 

above string represents the following trie/tree (visualize) and assume that there exisits a serialize method that performs above. 

Now, write a deserialize method so that above string to an object model of the following 

TreeNode 

TreeNode[] children

题目：给定以上的字典树序列化方法，请写出相应的反序列化方法。

解法：观察序列化的字符串，可以发现序列化的规则是“字母<若干个子树>”。用一个栈就可以写出比较简洁的反序列化代码。

代码：

 1 // http://www.careercup.com/question?id=5799446021406720

 2 #include <iostream>

 3 #include <stack>

 4 #include <string>

 5 #include <vector>

 6 using namespace std;

 7 

 8 struct TrieNode {

 9     char ch;

10     vector<TrieNode *> child;

11     TrieNode(char _ch = 0): ch(_ch), child(vector<TrieNode *>()) {};

12 };

13 

14 TrieNode *deserializeFromString(const string &str)

15 {

16     TrieNode *root;

17     TrieNode *ptr;

18     stack<TrieNode *>  st;

19     

20     int i, n;

21     

22     root = nullptr;

23     n = (int)str.length();

24     i = 0;

25     while (i < n) {

26         if (str[i] == '>') {

27             st.pop();

28             ++i;

29         } else {

30             ptr = new TrieNode(str[i]);

31             i += 2;

32             if (st.empty()) {

33                 root = ptr;

34             } else {

35                 (st.top()->child).push_back(ptr);

36             }

37             st.push(ptr);

38         }

39     }

40     

41     return root;

42 }

43 

44 void preorderTraversal(TrieNode *root)

45 {

46     if (root == nullptr) {

47         return;

48     }

49     cout << root->ch << ' ';

50     for (int i = 0; i < (int)root->child.size(); ++i) {

51         preorderTraversal(root->child[i]);

52     }

53 }

54 

55 int main()

56 {

57     TrieNode *root = nullptr;

58     string str;

59     

60     while (cin >> str) {

61         root = deserializeFromString(str);

62         preorderTraversal(root);

63         cout << endl;

64     }

65     

66     return 0;

67 }